I want to execute my shell with the environment I provide as the command line arguments. For that I have a script that ends with exec zsh -d -f after setting all the variables I want, which gives me a new shell with all variables set.
e.g:
export MY_SESSION="$1"
cd $2
export PS1="$3; "
exec zsh -d -f
My issue is that, I also want to execute bindkey -e on the new shell before it is made available. How do I do that?
I managed to do this with expect. Here is what I did.
#!/usr/bin/env expect
spawn /bin/zsh -d -f
send "source ./bin/activate\r"
send "export PS1='*| '\r"
send "bindkey -e\r"
interact
This successfully sends the bindkey -e on the produced zsh shell.
Related
On mac OSX, I have this script:
#!/usr/local/bin/bash
echo -e "\e[41mError: Some error.\e[0m"
When I just run the echo -e ... in a console, it prints the colored text "Error: Some error."
When executed as a script sh myscript.sh, it litterally prints the flag and the escape characters: -e "\e[41mError: Some error.\e[0m".
When I add the script location to ~/.bash_profile and execute it as myscript.sh, it does work. But I need to be able execute it without adding it to my bash profile.
Edit: using printf works: printf "\e[41mError: Some error.\e[0m\n".
when you run the shell with sh it runs in posix compatibility mode (i.e. as the bourne shell does)
bash is a successor to this shell, one of the features it adds is the -e switch to echo
in posix shell you don't need the -e, the escapes will be evaluated anyway
in bash you do, so if you want to run bash do so explicitly
I am attempting to store the result of an echo command as a variable to be used in a shell script. Debian 4.19.0-6-amd64
The command works in terminal: echo $HOSTNAME returns debian-base, the correct hostname.
I attempt to run it in a shell script, such as:
#!/usr/bin/bash
CURRENT_HOSTNAME=`echo $HOSTNAME`
echo $CURRENT_HOSTNAME
I have tried expansion:
CURRENT_HOSTNAME=$(echo $HOSTNAME)
And just to cover some more bases, I tried things like:
CURRENT_HOSTNAME=$HOSTNAME
# or
CURRENT_HOSTNAME="$HOSTNAME"
# also, in case a problem with reserved names:
test=$HOSTNAME
test="$HOSTNAME"
Works great in the terminal! Output is as follows:
root#debian-base:/scripts# echo $HOSTNAME
debian-base
root#debian-base:/scripts# TEST_HOSTNAME=$HOSTNAME
root#debian-base:/scripts# echo $TEST_HOSTNAME
debian-base
root#debian-base:/scripts# TEST_TWO_HOSTNAME=$(echo $HOSTNAME)
root#debian-base:/scripts# echo $TEST_TWO_HOSTNAME
debian-base
As soon as I run the script (as above):
root#debian-base:/scripts# sh test.sh
root#debian-base:/scripts#
What am I doing wrong?
You are using bash as your terminal. Bash has the variable $HOSTNAME set. You run your script with sh. sh does not have a $HOSTNAME.
Options:
bash test.sh
Or run it as a program:
chmod +x test.sh
./test.sh
But I think you need to change your first line to:
#!/bin/bash
As I don't think bash is installed in /usr/bin in most cases. But you need to try. To figure out where bash is installed use which bash
Another option is to use the hostname binary:
CURRENT_HOSTNAME=$(hostname)
echo $CURRENT_HOSTNAME
Which works in both bash and sh.
You can start sh by just running sh. You will see it has a bash-like terminal. You can try to do echo $HOSTNAME. It will not show, because it's not there. You can use set to see all the variables that are there (as sh does not have tab completion it's harder to figure out).
I'm trying to see what the output of a command would be if I were in a login shell, without having to go into a login shell. I've tried several variations of
zsh --login -c "alias"
But none of my aliases get shown; are --login and -c incompatible?
To test the difference between zsh --login -c "alias" and a normal login shell, you can/should add the -x option to see what the shell is up to.
When I run zsh -x --login -c "alias", then it processes /etc/zprofile.
When I run zsh -x --login, then it processes /etc/zprofile and /etc/zshrc.
I don't normally use zsh, so I don't have any personalized profile or start up file for it, but it seems plausible that it might look for (but, in my case, not find) ~/.zprofile and ~/.zshrc too.
I created trivial versions of those files:
$ echo "echo in .zprofile" > ~/.zprofile
$ echo "echo in .zshrc" > ~/.zshrc
and sure enough, they're processed. Further, the -c command with --login processed the .zprofile but did not process the .zshrc file.
Thus, using -c "alias" after the --login suppresses the processing of /etc/zshrc and ~/.zshrc. If you want those executed even so, you need to use something like:
zsh --login -c "[ -f /etc/zshrc ] && . /etc/zshrc; [ -f ~/.zshrc ] && . ~/.zshrc; alias"
Using -x to debug login processing is often informative.
It's nice that modern shells provide a command line option to induce login processing. I still have a program (which I don't use any more) that runs a login shell the old-fashioned way, by adding a - before the shell name in argv[0]. Thus, running -ksh would trigger login processing; the login program would run the login shell with the - at the start.
I'm new to bash and trying to understand what the script below is doing, i know -e is exit but i'm not sure what -se or what the $delimiter is for?
$delimiter = 'EOF-MY-APP';
$process = new SSH(
"ssh $target 'bash -se' << \\$delimiter".PHP_EOL
.'set -e'.PHP_EOL
.$command.PHP_EOL
.$delimiter
);
The -s options is usually used along with the curl $script_url | bash pattern. For example,
curl -L https://chef.io/chef/install.sh | sudo bash -s -- -P chefdk
-s makes bash read commands (the "install.sh" code as downloaded by "curl") from stdin, and accept positional parameters nonetheless.
-- lets bash treat everything which follows as positional parameters instead of options.
bash will set the variables $1 and $2 of the "install.sh" code to -P and to chefdk, respectively.
Reference: https://www.experts-exchange.com/questions/28671064/what-is-the-role-of-bash-s.html
From man bash:
-s If the -s option is present, or if no arguments remain after
option processing, then commands are read from the standard
input. This option allows the positional parameters to be
set when invoking an interactive shell.
From help set:
-e Exit immediately if a command exits with a non-zero status.
So, this tells bash to read the script to execute from Standard Input, and to exit immediately if any command in the script (from stdin) fails.
The delimiter is used to mark the start and end of the script. This is called a Here Document or a heredoc.
This should be very simple.
I recently noticed that when I type 'bash' into Terminal on Mac it shows this:
Jays-MacBook-Pro: ~ $ bash
bash: parse_git_branch: command not found
When before it didn't. Can someone explain why and how to resolve.
It is likely that you configured BASH to run parse_git_branch and print the result as part of PS1 (or alike). You can check this by: "echo $PS1" and "echo $PROMPT_COMMAND".
However, parse_git_branch is not a builtin function of bash. Below is how I configured my PS1. You may want to copy my git_branch_4_ps1 as your parse_git_branch
PS1='\n' # begin with a newline
PS1=$PS1'\[\e[38;5;101m\]\! \t ' # time and command history number
PS1=$PS1'\[\e[38;5;106m\]\u#\h ' # user#host
PS1=$PS1'\[\e[7;35m\]${MY_WARN}\[\e[0m\] ' # warning message if there is any
PS1=$PS1'\[\e[38;5;10m\]${MY_EXTRA} ' # extra info if there is any
PS1=$PS1'\[\e[0;36m\]$(git_branch_4_ps1) ' # git_branch_4_ps1 defined below
PS1=$PS1'\[\e[38;5;33m\]\w' # working directory
PS1=$PS1'\n\[\e[32m\]\$ ' # "$"/"#" sign on a new line
PS1=$PS1'\[\e[0m\]' # restore to default color
function git_branch_4_ps1 { # get git branch of pwd
local branch="$(git branch 2>/dev/null | grep "\*" | colrm 1 2)"
if [ -n "$branch" ]; then
echo "(git: $branch)"
fi
}
If your parse_git_branch is defined in ~/.bash_profile, it will not be loaded when you open a non-login shell (e.g. by running bash).
The differences between login and non-login shells are described here: Difference between Login Shell and Non-Login Shell? For our purposes, the main difference is that login shells (e.g. that when you first open Terminal) automatically source ~/.bash_profile upon startup, whereas non-login shells (e.g. that when you run bash from within Terminal) do not.
To fix this error, simply source your ~/.bash_profile after running bash:
user#host:~ $ bash
bash: parse_git_branch: command not found
user#host:~ $ source .bash_profile
Alternatively, place the function in ~/.bashrc instead, which will be automatically sourced by non-login shells (as covered in the earlier link).
Instead of having
parse_git_branch
call in PS1 definition alone you may use
parse_git_branch 2>/dev/null
to send stderr to /dev/null. This will silence the error you don't want to see.
have you export your $PS1 ?
You can check by run command:
printenv
else you should export it by run:
export -n PS1
after you will can run sudo or sudo su without problem
The key to this is to NOT export PS1. If it's exported, then any non-login shell also takes PS1. Since .bash_profile is automatically source'd by the login shell, the PS1 variable only affects the login shell.